SOLUTION OF THE EQUATIONS 
131 
It now appears very unlikely that any of the stars, except probably 
the mysterious “white dwarfs,” deviate appreciably from a perfect gas, 
so that these early attempts to treat the dwarf stars have lost interest. 
But we cannot ignore the study of imperfect gases entirely. It is part of 
our task to set forth the astronomical evidence that leads to the conclusion 
that the dwarf stars obey the gas laws, and for this it is necessary to discuss 
what would happen if they did not. 
Van der Waals’ equation for an imperfect gas is 
the volume v being the reciprocal of the density. The term a/v 2 can be 
neglected in comparison with the enormous pressures in the stars. The 
equation can then be written 
and p 0 represents the limiting density when the pressure is infinite. 
It would be idle to discuss the validity of this equation in actual stellar 
conditions because our ultimate conclusion will be that it does not apply 
at all, the supposed deviations due to the finite size of the atoms having 
no existence. We take (95*1) as giving a model star deviating from a 
perfect gas in a way similar to terrestrial gas. 
It is assumed as before that rjk is constant. 
The results (84*1), (84-2) and (84-3) apply if p is replaced by p '; and 
the equations of equilibrium are accordingly 
We choose p 0 according to the extent of the deviation from a perfect 
gas which we wish to consider. We then start at the centre of the star 
with arbitrarily chosen values of k and central density p c and build up 
a solution of (95-3) by quadratures. The mass M and radius R are found 
at the end of the quadratures, when p has been brought down by steps to 
zero. Unlike the problem of the point-source (§91) the solution presents 
no difficulty and the quadrature is of the simplest kind. 
If in any solution of (95*3) all masses are altered in the ratio M and 
all lengths in the ratio M % densities will be unaltered and the left side of 
the equation is unaltered. Since gdr is then altered in the ratio M%, k 
must be altered in the ratio MK 
(V + a/^ 2 ) (v — b) = 9IT/p,, 
.(95-1), 
.(95-2), 
where 
P = P (1 ~ P/Po) 1 
dP 
P = K P '\ 
- d (p'i) =- g dr. 
Hence 
P x 
The left side can be integrated, giving 
d {4/>'* (1 + p'/4p 0 )} = - (g/ K ) dr 
(95-3).